Learning Combined Equation

Learning Combined equation deals with an equation in which a homogeneous function is equal to zero. Combined equation relates with the Homogeneous function in which through the equation, in the given way as like the sum of the indices occurring in each term is constant. For e.g. x³ + 2x² + y³ +5xy², here degree of each term is 3. by this way learning of combined equation is proceeded.

 

Learning Combined equation – Definition

 

A combined equation of the two lines all the way through the origin is a homogenous equation of second degree. – Let the two strokes overtake through the starting point be y = m 1 x and y = m 2 x. i.e. y – m 1 x = 0 and y – m 2 x = 0 their combined equation then this is clearly a homogeneous or combined equation… ²A normal, integral, algebraic equation of the variables x and y is said to be homogeneous equation of n^th degree in x and y, when the sum of the indices of x and y in every term is the same and is equal to n.

I am planning to write more post on Inverse Functions Calculus with example, How to Solve Multi Step Inequalities. Keep checking my blog.

 

Example for learning combined equation

 

Combined equations of two outlines are in all the way through the origin as a homogeneous equation of the second degree

Let,

y = S₁x … (1)

y = S₂x … (2)

Be the lines through the origin.

The equations (1) and (2) can be written as

y – S₁ x = 0 and y – S₂x = 0

Their combined equation is

(y – S₁X)(Y- S2X) = 0

Y² – (S1 + S2)XY + S1S2X² = 0  …(3)

This is clearly a homogenous equation of the second degree in x and y.

To express ax² + 2hxy + by² = 0 as bS² + 2hS + a =0

ax² + 2hxy + by² = 0

Divide equation (1) by x²,

a+ 2h`(y)/(x)` + b(`(y)/(x)`)² = 0

Substituting `(y)/(x)` = S, we get

a+ 2hS + bS² = 0

This can be written as bS² + 2hS + a = 0

This is a quadratic form in m. This has two roots (say) S₁ and S_2.

Therefore S₁ + S₂  = `(-2h)/(b)` ,  S₁S₂ = `(a)/(b)`

 

2nd Grade Algebra

Introduction :

Algebra is the branch of mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures. Together with geometry, analysis, topology, combinatory, and number theory, algebra is one of the main branches of pure mathematics. (Source: Wikipedia)
Example Problems for 2nd Grade Algebra:

2nd grade math – Addition problem:

There are 12 boxes in the store. We bought 10 more boxes for the store. How many boxes are there in the store now?

Solution:

In 12 boxes in the store

And then we bought 10 boxes in the store

12 + 10 = 22

Therefore 22 boxes in the store now.

2nd grade math -Subtraction problem:

A vegetable seller had 15 onions. He sold 6 onions. How many onions did he have left?

Solution:

The total amount of vegetables = 15 onions.

Sold onions = 6

Remaining onions =?

So, 15 – 6 = 31

He had 9 onions left.

2nd grade math – Multiplication problem:

There are 5 apples in each box. How many are there in 4 boxes?

Solution:

So, 5 × 4 = 20

There are 20 apples in 4 boxes.

2nd grade math – Division problem:

Peter bought a sack of 18 kg of flour. He has packed the flour equally into the 3 bags. How many kilograms of flour were there in each bag?

Solution:

18 ÷ 3 = 6

There were 6 kg of flour in each bag.
Algebra is widely used in day to day activities watch out for my forthcoming posts on algebra 1 and free online math tutor. I am sure they will be helpful.

Practice Problems for 2nd Grade Algebra:

1.There are 10 boxes in the store. We bought 7 more boxes for the store. How many boxes are there in the store now?

[Answer: 3]

2. A fruit seller had 10 bananas. He sold 2 bananas. How many bananas did he have left?

[Answer: 8]

3. There are 4 oranges in each box. How many are there in 2 boxes?

[Answer: 8]

4. Jhon bought a sack of 20 kg of flour. He has packed the flour equally into the 2 bags. How many kilograms of flour were there in each bag?

[Answer: 10]

Linear Equations with Fractions Calculator

Introduction :

A statement of equality which contains one or more unknown quantity or variable is called an equation. An equation involving only linear polynomials is called a linear equation. A linear equation having fractions is known as linear equation with fractions.

Rules for solving linear equation with fractions calculator:

Same number can be added to both sides of an equation without changing the equality
Same number can be subtracted from both sides of an equation without changing the equality
Both sides of an equation may be multiplied by the same non-zero number without changing the equality.
Both sides of an equation may be divided by the same non-zero number without changing the equality.

Examples to Solve Linear Equtions with Fractions Calculator

1) `x/5 + 11 = 1/5`

Solution: `x/5 + 11 = 1/5`

subtracting 11 on both sides

`x/5+11-11 = 1/5 -11`

`x/5 = 1/5 – 11`

Least common denominator is 5,

`x/5 = 1/5 – (11*5)/5`

`x/5 = (1-55)/5`

`x/5 = (-54)/5`

multiply 5 on both sides.

`x/5*5 = (-54)/5 * 5`

x = -54

2) `x/3-5/2 = 6`

Solution: we have, `x/3-5/2 = 6`

LCD = 2*3 = 6

`(x*2)/(3*2) – (5*3)/(2*3) = 6`

`(2x+15)/6 = 6`

multiply throughout by 6 on both sides

`(2x+15)/6 * 6 = 6*6`

2x+15 = 36

subtract 15 on both sides,

2x + 15 – 15 = 36 – 15

2x = 21

Divide by 2 on both sides

`(2x)/2 = 21/2`

` x = 21/2`

Algebra is widely used in day to day activities watch out for my forthcoming posts on Adding Fractions with Variables and Ordering Fractions. I am sure they will be helpful

More Examples to Solve Linear Equtions with Fractions Calculator

1) `(2x-1)/3 – (6x-2)/5 = 1/3`

Solution: Given, `(2x-1)/3 – (6x-2)/5 = 1/3`

LCD = 3 * 5 = 15

`((2x-1)*5)/(3*5) – ((6x-2)*3)/(5*3) = (1*5)/(3*5)`

`(((2x-1)*5)-((6x-2)*3))/15 = 5/15`

(2x-1)* 5 – (6x-2)*3 = 5

10x -5 -18x +6 = 5

10x-18x -5+6 = 5

-8x +1 = 5

subtract 1 on both sides

-8x +1-1 = 5-1

-8x = 4

divide by (-8) on both sides

`(-8x)/(-8) = 4/(-8)`

`x = (-1/2)`

2) `(17(2-x)-5(x+12))/(1-7x) = 8`

Solution: we have,

`(17(2-x)-5(x+12))/(1-7x) = 8`

`(34-17x -5x -60)/(1-7x) = 8`

`(-22x-26)/(1-7x) = 8`

multiply throughout by (1-7x)

`(-22x-26)/(1-7x)*(1-7x) = 8*(1-7x)`

-22x -26 = 8(1-7x)

-22x -26 = 8-56x

add 56x on both sides

-22x – 26 +56x = 8-56x+56x

-22x+56x -26 = 8

34x -26 = 8

add 26 on both sides

34x -26+26 = 8+26

34x = 34

divide 34 on both sides

`(34x)/34 = 34/34`

x = 1

Quality Differentiation

The derivative, events the price at which the needy variable changes with revere to the independent variable. It is single of the most significant ideas in Calculus. The quality differentiations of functions are extensively used in science, finances, medicine and computer science. Additional split the class of continuous functions into two sub classes, derivable and non-derivable. After having continuous functions, limits and stability, we shall further split the group of continuous functions into two sub classes, derivable and non-derivable.
Formula for Quality Differentiation

f'(x) = Lim f(x+∂x) – f(x)

∂x → 0 ——————–

∂x

This equation is well known as quality differentiation equation. By use this equation we preserve get derivatives of dissimilar functions. ∂x is the little modify in the variable x. f(x) is a purpose of x. f ‘(x) is derivative of f(x).
Example Problems for Quality Differentiation

Ex: 1 Find the derivative of 8x + 10

Sol:

Let f(x) = 8x + 10 then f(x + ∂x) = 8(x + ∂x) + 10

f'(x) = Lim f(x+∂x) – f(x)

∂x → 0 ——————–

∂x

= Lim 8(x + ∂x) + 10 – (8x + 10)

∂x → 0 —————————

∂x

= Lim 8x + 8∂x + 10 – 8x -10

∂x → 0 ————————

∂x

= Lim 8

∂x → 0

= 8.

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Ex 2: Find the derivative of 5x + 20

Sol:

Let f(x) = 5x + 20 then f(x + ∂x) = 5(x + ∂x) + 20

f'(x) = Lim f(x+∂x) – f(x)

∂x → 0 ——————–

∂x

= Lim 5(x + ∂x) + 20 – (5x + 20)

∂x → 0 —————————

∂x

= Lim 5x + 5∂x + 20 – 5x -20

∂x → 0 ————————

∂x

= Lim 5

∂x → 0

= 5.

Ex 3: Find the derivative of the equations f(x) = 3 + 5cos x, find f ‘(x).

Sol:

f'(x) = d(3 + 5cos x)

————–

dx

= d(3) + d(5cos x)

—- ———

dx dx

= 0 + 5(-sin x)

= -5sin x.

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Inverse proportions

Introduction:

Two quantities ‘a’ and ‘b’ are said to vary inversely to each other if the product ‘ab’ always remains constant.  The product is known as constant of proportionality.

Let ‘a’  and ‘b’ be two quantities, if ab = k,  where k is a constant of proportionality, then ‘a’ and ‘b’ are said to be inversely proportional to each other.

Example: If ‘a’ and ‘b’ vary inversely, then find the constant of proportionality if a = 8 and b = 10

Solution:  We know, if ‘a’ and ‘b’ are inversely proportional then,

ab = k,               where k is a constant

Here, given a = 8 and b = 10

Therefore,  ab = 8 * 10 = 80

So, k = 80

Examples of Inversely Proportional

1) If x and y vary inversely as each other.  x = 10 when y = 6. Find y when x = 15

Solution: Given x and y are inversely proportional,

Therefore,      xy = k                  where k is a constant

For   x = 10  and y = 6,

xy   =  k

10 * 6  = k

60 = k

Now, for x = 15, y = ?

xy = 60

15 * y = 60

y  = `(60)/(15)`

y  =  4

2) If x and y vary inversely as each other and x = 5 when y = 15.  Find x when y = 12

Solution: Given x and y are inversely proportional,

Therefore,  xy  =  k              where k is a constant,

For  x = 5 and y = 15

x y = k

5 * 15 = k

75 = k

For y = 12,  x = ?

x * 12 = 75

x =   `(75)/(12)`  =  `(25 * 3)/(4 * 3)`  =  `(25)/(4)`

x = 6.25

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3) If x and y vary inversely as each other. If  x = 30, find y when constant of proportionality = 900

Solution: Given x and y are inversely proportional,

so,  xy  = k             where k is constant,

For  x = 30,  k = 900,  y = ?

30 * y = 900

y =  `(900)/(30)`  = `(30 * 30)/(30)`  = 30

y  = 30

4)  If x and y vary inversely as each other and if y = 35 find x when constant of proportionality is 7

Solution: Given x and y are inversely proportional,

Therefore,    xy = k,               where k is constant

For  y = 35,  k = 7 , x = ?

x * 35 = 7

x  =   `(7)/(35)`  =   `(7)/(7*5)`  =   `(1)/(5)`

x  =  `(1)/(5)`

Word Problems on Inversely Proportional

1) If 52 men can do a piece of work in 35 days, in how many days 28 men will do it?

Solution: Let 28 men do the work in x days,

Clearly, less the number of men,  more will be the number of days to finish the work. So, this case is an inverse proportional.

Ratio of number of men   =  Inverse ratio of number of days.

`=>`                         52 : 28     =    x : 35

`=>`                                  `(52)/(28) = (x)/(35)`

`=>`                            52 * 35  =  28 * x                      cross multiply the terms

`=>`                                       x  =  `(52* 35)/(28)`

`=>`                                           =  65

Hence, 28 men will do the work in 65 days.

2)  Maria cycles to her school at an average speed of 12 km/hr. It takes her 20 minutes to reach the school. If she wants to reach her school in 15 minutes, what should be her average speed?

Solution:  Let the required speed be x.

Clearly, less the number of men, more will be the number of days to finish the work. So, this case is an inverse proportional

Ratio of speeds  =  Inverse ratio of time taken

`=>`                         12 : x     =   15 : 20

`=>`                               `(12)/(x) = (15)/(20)`

`=>`                            15  * x  =  12  * 20                                 cross multiply the terms

`=>`                                       x  =  `(12* 20)/(15)`

`=>`                                           =  16

Hence, Maria average speed should be 16 km/hr.

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Proportion Word Problems by commenting on blog

To practice algebra linear equation

Introduction :

Algebra linear equation is a mathematical expression that contains an equal sign and the linear expressions. Algebra is a Arabic word al–jabr.  In mathematics, we use letters like a, b, x and y to denote numbers. We performs addition, subtraction, multiplication, division or extraction of roots  and real numbers, we obtain what are called algebraic expressions. Symbols in an algebraic expression are called variables . An  Algebraic expression of the form axⁿ is called a monomial in x,

Practice Algebra Linear Equation – Example

A set of common solution set of practice algebraic expression equation are said to be linear equations. A linear equations are set of variables in equation form

For example, this linear equation:  x + 1 = 4 means that when we add 1 to the unknown value, ‘x’, the answer is equal to 4.

Two variable equation: 2x+3y=6 where x and y are two variables.

Practice Algebra Linear Equation – Solved Examples

 Example 1:

Solve for x

           X+ 2 =   -3

solution:

1. Subtract 1 from both sides:

x + 2 – 1 =   -3 – 2

2. Simplify both sides:

x   =   -5

Example 2:

Solve for x

-3x   =   12

solution:

1. Divide both sides by -3:

`(-3x)/-3` =`12/-3`

2. Simplify both sides:

          x   =   -4

Example 3:

Solve for x

`x/4` =-2

solution:

1. Multiply both sides by 4:

X*4/4=-2*4

2. Simplify both sides:

x   =   -8

Example 4:

Solve for x

2 x + 2 = -18

solution:

1. Subtract 1 from both sides:

2x + 2 -2 = -18 – 2

2. Simplify both sides:

2x = -20

3. Divide both sides by 2:

2x/2=-20/2

4. Simplify both sides:

x   =   -10

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Example 5:

Solve the given Equation: 8z + 50 = 4z + 62

Solution:

Given equation:    8z + 50 = 4z + 62

Subtracting both sides from 50:    8z + 50- 50 = 4z + 62 – 50

Simplification gives:        8z = 4z + 12

Subtracting both sides 4z:  8z – 4z = 4z – 4z + 12

Simplification gives:      4z = 12

Answer:    z = 3

Example 6:

         Given question is, Y = 3x +4

Solution: General equation is y = mx + b

Where m = Slope of the line

b = y- intercept

(Slope) m= 3

Y-intercept = 1

Here the values of slope (3) and intercept (1) are the prepared values from the given linear equation.

Practice Algebra Linear Equation – Practice Problems:

Problem 1:

       solve for x in (x+5)=7

Solution:

       x=-2

 

Problem 2:

      solve `x/5` +3=5

Solution:

      x= -10

Problem 3:

      solve x+3x-5=4

Solution:

       x= `9/4`

Problem 4:

      solve 2(17x+5)=2

Solution:

      x = -`5/17` .